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Timeline for Different definitions of category

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Apr 14, 2018 at 8:39 comment added Philippe Gaucher "Because for many important categories, the most natural presentation gives non-disjoint homsets." (like for the category of sets). No it is wrong at least for me :-). A set map consists of a triple $(X,Y,f)$ where $f$ is a subset of $X\times Y$ such that $(x,y)\in f$ and $(x,y')\in f$ implies $y=y'$.
Apr 12, 2018 at 22:26 comment added Peter LeFanu Lumsdaine @JohannesHahn: Nice example, but to guard against misinterpretation let me add the caveat that again the difference is just cosmetic. There's no serious obstruction to defining the functor, one just has to make sure one defines functor a in a way that matches the arbitrary-homsets definition of categories, by saying "for all objects x, y, a map hom(x,y) —> hom(Fx,Fy)…”
Apr 12, 2018 at 21:45 comment added Johannes Hahn Concrete example: $X=Y=S^1, Z=\mathbb{R}^2$ and the inclusion/identity map. After applying the functor $H_1$ ($H^1$ wuld work equally well) the identity map becomes the identity $\mathbb{Z}\to\mathbb{Z}$, the inclusion map becomes the zero mapping $\mathbb{Z}\to\{0\}$.
Apr 12, 2018 at 21:44 comment added Johannes Hahn Having disjoint homsets is also necessary to make for nicer definitions of functors. For example, if one has a concrete category, say Top, and encodes morphisms non-disjointly, say by identifying a continuous function with its graph, you get well-definedness issues: The same graph may describe morphisms $X\to Y$ and $X\to Z$ where $Y\subseteq Z$, but after applying a functor like (co)homology, $H(Y)$ and $H(Z)$ do not need to have an inclusion between them so that the image of two morphisms is now not compatible any more in the same way.
Apr 12, 2018 at 20:09 vote accept Mahdi Majidi-Zolbanin
Apr 12, 2018 at 20:04 history answered Peter LeFanu Lumsdaine CC BY-SA 3.0